For all matrices A exists unique matrix B with $B^2=A$ in a sufficient neighborhood.

Show existence of $$\epsilon > 0, \delta > 0$$ such that for all $$A\in U_\epsilon(Id_n)$$ exists unique $$B \in U_\delta(Id_n)$$ with $$B^2=A$$.

$$U_\epsilon(Id_n):=\{X \in M_{n\times n}| \Vert X-Id_n \Vert < \epsilon; \Vert \cdot \Vert \text{ Operator norm}\}$$

Implicit function method: Define $$F:(A,B) \mapsto B^2-A$$ and noticing that $$(Id_n,Id_n)$$ is a solution, and also $$\partial_{B}F|_{(Id_n,Id_n)}=2Id_n \in GL(n)$$. Impl.f.th. guarantees an existence of unique function $$g:U_\epsilon(Id_n)\to U_\delta(Id_n)$$ such that $$g(A)^2-A=0$$. Which is equivalent to the statement if we set $$B:=g(A)$$.

Now I would like to show same using inverse function theorem. Define $$f(B)=B^2$$, then its differential is $$df_{Id_n}=2Id_n$$ invertible. Therefore we get open neighborhoods $$U,V$$ of $$Id_n$$ and $$f(Id_n)=Id_n$$ respectively with $$f|_{U}:U\to V$$ diffeomorphism.
In particular, bijectivity implies $$\forall A\in V: \exists! B \in U: f(B)=B^2=A$$

Question: How to shrink $$U,V$$ to the form of open balls?

Since $$V$$ is open, you can find a ball $$B_1$$ of radius $$\delta$$ around the identity that is contained in $$V$$. Now, look at the preimage $$X=f^{-1}(B_1)\subseteq U$$. It is open, since $$f$$ is continuous. Therefore you can find a ball $$B_2$$ of radius $$\epsilon$$ around the identity that is contained into $$f^{-1}(B_1)$$ therefore into $$U$$. These should do the trick.
PS: I've used here that the open balls form a basis of your topology around $$Id_n$$.